Question

When electron in hydrogen atom jumps from second orbit to first orbit, the wavelength of radiation emitted is $\lambda$. When electron jumps from third orbit to first orbit, the wavelength of emitted radiation would be

• A. $\frac{27}{32}\lambda$
• B. $\frac{32}{27}\lambda$
• C. $\frac{2}{3}\lambda$
• D. $\frac{3}{2}\lambda$

Answer 1

The correct option is A $\frac{27}{32}\lambda$

Wavelength of photon emitted $\frac{1}{\lambda }=R(\frac{1}{n_2^2}-\frac{1}{n_1^2})$ $(\because Z=1)$

Transaction from second orbit to first orbit :

We have, $n_1=2$ $n_2=1$

$\therefore$ $\frac{1}{\lambda }=R(\frac{1}{1^2}-\frac{1}{2^2})$ $⟹R=\frac{4}{3\lambda }$

Transaction from third orbit to first orbit :

We have, $n_1=3$ $n_2=1$

$\therefore$ $\frac{1}{{\lambda }^{\prime }}=R(\frac{1}{1^2}-\frac{1}{3^2})$

$⟹{\lambda }^{\prime }=\frac{9}{8R}=\frac{9}{8}\times \frac{3\lambda }{4}=\frac{27}{32}\lambda$